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# Hybridisation and shape of $Xe{F_4}$ is:(A) $s{p^3}d$, Trigonal bipyramidal(B) $s{p^3}$, Tetrahedral(C) $s{p^3}{d^2}$, Square planar(D) $s{p^3}{d^2}$, Hexagonal

Last updated date: 23rd Jul 2024
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Hint: To determine the shape and the geometry of the molecule, we use VSEPR theory. According to VSEPR theory, the force of repulsion between the lone pairs of electrons on the central atom should be minimal. This is to stabilize the molecule.

Let’s solve this question using VSEPR theory and the valence shell of the xenon, $Xe$ .
Since, we know that our central atom is xenon, $Xe$ and it has $8$ valence electron in the valence shell, and out of these $8$ electrons, $6$ electrons are in the $5p$ orbital and the $2$ electrons are in the $5s$ orbital.
Also, the $5d$ and the $5f$ orbitals of the xenon are empty. So during the formation of $Xe{F_4}$ , the two excited electrons of the $5p$ orbital move to the empty $5d$ orbital. As a result, there are $4$ unpaired electrons, $2$ in the $5p$ orbital and $2$ in the $5d$ orbital.
This all over arrangement of electrons in the s,p,d orbitals gives $s{p^3}{d^2}$ hybridisation of the molecule.
And according to the VSEPR theory, the molecule with $s{p^3}{d^2}$ hybridisation has square planar geometry.
So, based on the above discussion, the correct option is (C). Therefore, Hybridisation and shape of $Xe{F_4}$ is $s{p^3}{d^2}$ and square planar respectively.
The four fluorine atoms pairs with the four half-filled orbitals (i.e. $5p$ and the $5d$ orbital) and they lie at the corners of the central atom to minimise the repulsion. Also, the two lone pairs of electrons of the central atom i.e. xenon lie perpendicular to the plane to make the molecule overall stable. Hence, this gives the molecule its square planar shape.